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The problem I've got says to differentiate the following:

$x = f(y) = y^2 + \frac 1y$

I am a student studying differentiation at the moment.

I don't quite get how am I supposed to differentiate $x$, as in $\frac{dx}{dx}$ or how?! Isn't the derivative of $x$ always $1$? As if we differentiate $y^2 + \frac 1y$, we get that the derivative of $x$ equals $2y - \frac{1}{y^2}$, which is clearly not 1... I am confused, it is either that the question is wrong or that I've missed something.

Would be helpful if someone can give me a hint about this. Thanks.

User 123
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zаѓатhᵾѕтѓа
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2 Answers2

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$\frac{dx}{dx} = 1$, $\frac{dy}{dy} = 1$ for any equation.

Your equation is $x = f(y) = y^2 + \frac 1y$

Note that $x$ is written in terms of $y$

You can differentiate $x$ in terms of $y$, i.e., $\frac{dx}{dy}$

$\frac{dx}{dy} = 2y -\frac{1}{y^2}$

Your claim that $\frac{dx}{dx} = 2y -\frac{1}{y^2}$ is wrong.

If you want to differentiate $y$ in terms of $x$, i.e., $\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{(\frac{dx}{dy})}$ , in some cases.

Else, you can write the equation as y in terms of x, and then differentiate it.

User 123
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$$x=f(y)=y2+1y$$ steps (Sorry for bad englisch): Rewrite $$x=f(y)=y2+1y$$ in the standard form so it will become this: $$4\cdot \frac{1}{4}\left(x-\left(-\frac{1}{4}\right)\right)=\left(y-\left(-\frac{1}{2}\right)\right)^2$$ Therefore parabole properties are: $$\left(h,\:k\right)=\left(-\frac{1}{4},\:-\frac{1}{2}\right),\:p=\frac{1}{4}$$ hope you like this answer $$:)$$

Maksym Mazura Igorevič
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